I have the following summation: $$\sum\limits_{k=7}^{n} {k-1\choose 6} $$ and apparently it counts the number of subsets of {1, 2, . . . , n} having size 7.
Why is this?
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You probably know that ${n\choose 7}$ is the number of subsets of $\{1,2,\ldots,n\}$ of size $7$.
Each such subset has a largest element. The largest element must be one of $7,8,\ldots,n$. For each possible largest element, let's count the number of subsets having that as a largest element. We already chose the largest, so there are $6$ elements left to choose. So we add all the subsets of size $7$ with the largest element $7$: choose $6$ elements from $\{1,2,3,4,5,6\}$, that is ${6\choose 6}$. Then the largest element $8$: that is ${7\choose 6}$, and so on.
The final count of the subsets is
$${6\choose 6}+{7\choose 6}+\cdots +{n-1\choose 6}$$ $$=\sum_{k=7}^{n} {k-1\choose 6}$$
which is therefore equal to
$${n\choose 7}$$
Rory Daulton
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1I like your solution better than mine. It gives intuition and is not overly mechanical (like mine). Kudos. – TravisJ Apr 11 '15 at 00:52
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@TravisJ: Thanks for the encouragement! Your answer is also nice: I like knowing both kinds of proof for combinatorial identities. – Rory Daulton Apr 11 '15 at 09:46
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It is because of Pascal's recurrence. The recurrence is that $\binom{i}{j}+\binom{i}{j-1} = \binom{i+1}{j}$ provided that $0< j\leq i$.
\begin{align*} \binom{6}{6}+\binom{7}{6} &= \binom{7}{7} + \binom{7}{6} \\ &= \binom{8}{7} \\ \binom{6}{6}+\binom{7}{6} + \binom{8}{6} &= \binom{8}{7} + \binom{8}{6} \\ &= \binom{9}{7} \\ \binom{6}{6} + \binom{7}{6} + \binom{8}{6} + \binom{9}{6} &= \binom{9}{7} + \binom{9}{6} \\ &= \binom{10}{7} \end{align*} This process repeats until you get to $\binom{n-1}{6}+\binom{n-1}{7}=\binom{n}{7}$. It helps to draw Pascal's Triangle and follow the additions down the tree...
TravisJ
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